JEE Main 2016 · Physics · Waves and Sound · easy

A sonometer wire stretched by a metal block hanging from its end vibrates in its fundamental mode with frequency 400 Hz. The block (density 8000 kg…

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A sonometer wire stretched by a metal block hanging from its end vibrates in its fundamental mode with frequency

400\ \mathrm{Hz}

. The block (density

8000\ \mathrm{kg\,m^{-3}}

) is now completely immersed in water of density

1000\ \mathrm{kg\,m^{-3}}

while the vibrating length remains the same. The new fundamental frequency of the wire is nearly

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Answer: (B) $374\ \mathrm{Hz}$

For a fixed length the fundamental frequency is

f=\dfrac{1}{2L}\sqrt{\dfrac T\mu}\propto\sqrt T

. In air the tension is

T=Mg

. When the block is fully immersed, buoyancy acts upward on it, so the tension becomes

T'=Mg-\rho_wVg=Mg\left(1-\dfrac{\rho_w}{\rho_b}\right)=Mg\left(1-\dfrac18\right)=\dfrac78Mg

. Hence

f'=f\sqrt{\dfrac78}=400\times0.9354=374\ \mathrm{Hz}

. Reducing the frequency linearly by

\tfrac18

gives

350\ \mathrm{Hz}

, but frequency varies as the square root of tension.

Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake